In this post I'd like to construct an extension to the Sleeping Beauty problem which is unsolvable from an thirder/SIA perspective. The core of the construction is having the number of wakings Beauty will experience follow a distribution with an infinite mean. This leads to Beauty being unable to assign normalized probability values when reasoning under SIA. Given this forum's fascination with anthropic reasoning and this problem, this may be of interest.

The original Sleeping Beauty problem

The following is a summary of the thirder/SIA approach to the original Sleeping Beauty problem. From what I can tell, this forum is very well versed here, so I won't go into too much background detail.

In the original problem, we have the following two steps for the experimenter:

Flip a coin. If heads, record the value $N = 0$ . If tails, record the value $N = 1$ .
Put Beauty to sleep [on Sunday]. She will be woken once on each of the following $N + 1$ days, with memory erased between wakings. (i.e. once if heads, twice if tails)

Let $M_{k}$ be the proposition "I am awake and today is the $k$ th day since the experiment began". (I.e. $M_{1}$ is "I am awake and it is Monday"; $M_{2}$ is "I am awake and it is Tuesday".) Also let $M$ be the proposition "I am awake", in the sense that $M = M_{1} \lor M_{2}$ .

The problem is then to give Beauty a value to assign to the conditional probability of heads given that she has been awoken, i.e. to

$P (N = 0 ∣ M)$ .

One version of the thirder/SIA calculation for this value then proceeds as follows.

$\begin{matrix} \frac{P (N = 0 ∣ M)}{P (N = 1 ∣ M)} & = \frac{P (M ∣ N = 0)}{P (M ∣ N = 1)} \frac{P (N = 0)}{P (N = 1)} = \frac{P (M_{1} ∣ N = 0) + P (M_{2} ∣ N = 0)}{P (M_{1} ∣ N = 1) + P (M_{2} ∣ N = 1)} \frac{P (N = 0)}{P (N = 1)} = \frac{P (M_{1} ∣ N = 0) + 0}{2 * P (M_{1} ∣ N = 1)} \frac{P (N = 0)}{P (N = 1)} = \frac{1}{2} \frac{P (N = 0 ∣ M_{1})}{P (N = 1 ∣ M_{1})} = \frac{1}{2} \frac{1 / 2}{1 / 2} = \frac{1}{2} . \end{matrix}$

Where the steps are respectively due to:

Bayes' law.
That $M = M_{1} \lor M_{2}$ and that $M_{1}$ , $M_{2}$ are disjoint propositions.
For the numerator, that $M_{2}$ is impossible given $N = 0$ (i.e. we cannot wake up on Tuesday if the coin came up heads). For the denominator, this is the part of the SIA assumption: given that the coin came up tails, and we cannot distinguish a waking on Monday or Tuesday, we must put equal weight on both possibilities.
Bayes' law.
Following SIA/thirder reasoning, we put equal weight on heads or tails given that we are waking on Monday. I leave the subtler point of justifying this assignment to the many expositions of thirder/SIA reason

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0naut

0naut